There is an alternative way of computing $\mathcal{L}\left\{\left\lfloor\dfrac{t}{a}\right\rfloor\right\}$.
Consider the sawtooth wave of period $a$. Let us define one period of this function by $f_a(t) = \dfrac{t}{a}(u(t) - u(t-a))$ (several periods of $f(t)$ are shown below).
We now can define $\left\lfloor\dfrac{t}{a}\right\rfloor = \dfrac{t}{a} - f(t)$ for any $t\geq 0$ (which results in the familiar graph below).
Therefore, $$\mathcal{L}\left\{\left\lfloor\frac{t}{a}\right\rfloor\right\} = \mathcal{L}\left\{\frac{t}{a}\right\} - \mathcal{L}\{f(t)\} = \frac{1}{as^2} - \mathcal{L}\{f(t)\}.$$
To compute the Laplace transform of $f(t)$, you'll need to recall the Laplace transform formula for periodic functions; that is, if $g(t)$ is piecewise continuous for $t\geq 0$ and is $p$-periodic, then
$$\mathcal{L}\{g(t)\} = \frac{1}{1-e^{-ps}}\mathcal{L}\{g_p(t)\} = \frac{1}{1-e^{-ps}}\int_0^p e^{-st}g_p(t)\,dt$$
where, again, $g_p(t)$ is one period of $g(t)$.
To get that formula, though, you'd have to rewrite the improper integral (from the integral definition) as an infinite series and do some straightforward computations. For now, let's assume that we have the luxury of using this formula. Also, to help us compute $\mathcal{L}\{f_a(t)\}$, recall the derivative formula $\mathcal{L}\{tf_a(t)\} = -\dfrac{d}{ds}\mathcal{L}\{f_a(t)\}$.
We now see that
$$\begin{aligned} \mathcal{L}\{f_a(t)\} &= \frac{1}{a}\mathcal{L}\{t(u(t) - u(t-a))\} \\ &= -\frac{1}{a}\frac{d}{ds}\left[\mathcal{L}\{u(t)\} - \mathcal{L}\{u(t-a)\}\right]\\ &= -\frac{1}{a}\frac{d}{ds}\left(\frac{1}{s} - \frac{e^{-as}}{s}\right)\\ &= -\frac{1}{a}\left(-\frac{1}{s^2} - \frac{-ase^{-as} - e^{-as}}{s^2}\right)\\ &= \frac{1-e^{-as}-ase^{-as}}{as^2}\end{aligned}$$
Thus,
$$\mathcal{L}\{f(t)\} = \frac{1}{1-e^{-as}} \mathcal{L}\{f_a(t)\} = \frac{1-e^{-as}-ase^{-as}}{as^2(1-e^{-as})}$$
and therefore, we have that
$$\mathcal{L}\left\{\left\lfloor\frac{t}{a} \right\rfloor\right\} = \mathcal{L}\left\{\frac{t}{a}\right\} - \mathcal{L}\{f(t)\} = \frac{1}{as^2} - \frac{1-e^{-as}-ase^{-as}}{as^2(1-e^{-as})} = \frac{e^{-as}}{s(1-e^{-as})}.$$
Consider the sum:
$\begin{align*}
S(s)
&= \sum_{n \ge 0} e^{-n s} \\
&= \frac{1}{1 - e^{-s}}
\end{align*}$
So your mystery sum is just:
$\begin{align*}
-S'(s)
&= \sum_{n \ge 0} n e^{- n s} \\
&= \frac{e^s}{(1 - e^s)^2}
\end{align*}$
The frobbing is valid as long as $e^{-s} < 1$, i.e., $s > 0$.
Best Answer
Since $f_1(t)\leqslant u(t)$ for all $t$ and $$\int_0^\infty |e^{-st}u(t)|\ \mathsf dt = \frac1s<\infty $$ for $s>0$, dominated convergence yields \begin{align} \mathcal L(f_1(t)) &= \int_0^\infty \left(e^{-st}\sum_{n=0}^\infty (-1)^n u(t-n)\right) \mathsf dt\\ &= \sum_{n=0}^\infty (-1)^n\int_n^\infty e^{-st}\ \mathsf dt\\ &= \sum_{n=0}^\infty (-1)^n \frac{e^{-ns}}s\\ &= \frac1s \sum_{n=0}^\infty (-e^{-s})^n\\ &= \frac1{s(1+e^{-s})}. \end{align} Since $f_2(t)\leqslant t$ for all $t$ and $$ \int_0^\infty |e^{-st}t|\ \mathsf dt = \frac1{s^2} $$ for $s>0$, dominated convergence yields \begin{align} \mathcal L(f_2(t)) &= \int_0^\infty \left(e^{-st}\sum_{n=0}^\infty u(t-n) \right)\mathsf dt\\ &= \sum_{n=0}^\infty \int_n^\infty e^{-st}\ \mathsf dt\\ &= \sum_{n=0}^\infty \frac{e^{-ns}}s\\ &= \frac1{s(1-e^{-s})}. \end{align} Since $f_3(T)\leqslant u(t)$ for all $t$ and $t\mapsto e^{-st}$ is integrable, by dominated convergence we have \begin{align} \mathcal L(f_3(t)) &= \int_0^\infty \left(e^{-st}\sum_{n=0}^\infty (t-n)(u(t-n)-u(t-n+1) \right)\mathsf dt\\ &= \sum_{n=0}^\infty \int_n^{n+1}e^{-st}(t-n)\ \mathsf dt\\ &= \sum_{n=0}^\infty \frac{\left(e^s-1-s\right) e^{-(n+1) s}}{s^2}\\ &= \frac{(e^s-1-s)e^{-s}}{s^2}\sum_{n=0}^\infty (e^{-s})^n\\ &= \frac{(e^s-1-s)e^{-s}}{s^2}\cdot \frac1{1-e^{-s}}\\ &= \frac{1+s-e^s}{s^2\left(1-e^s\right) }. \end{align}